Let the required equation be y = mx + c; where y = 1, m = f'(0), x = 0
f(a) = sec(a)
f'(a) = sec(a)tan(a)
f'(0) = sec(0)tan(0) = 0
y = mx + c
1 = 0(0) + c
c = 1
Therefore, required equation is
y = 1.
Answer:
More informally: The two's complement of an integer is exactly the same thing as its negation. ... It means "to find the negation of a number (i.e., its two's complement) you flip every bit then add 1"
4 1/5 + B = 9 3/5
21/5 + B = 48/5
B = 27/5
B = 5 2/5
Letter C is the correct answer!
Hope this helps! ;)
Answer:
The area of the region between the graph of the given function and the x-axis = 25,351 units²
Step-by-step explanation:
Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15
If 'f' is a continuous on [a ,b] then the function

By using integration formula

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

<em>On integration , we get</em>
= 

= 
After simplification and cancellation we get
= 
on calculation , we get
= 
On L.C.M 15
= 
= 25 351.2 units²
<u><em>Conclusion</em></u>:-
<em>The area of the region between the graph of the given function and the x-axis = 25,351 units²</em>
Answer:
can you show me the data so i can see the temperature